Why, mathematicians do it too. They just prefer to understand what exactly they're doing before doing it.
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Grigory MJun 23 '11 at 7:55

1

Could you please give an example of a derivation in your textbook upon which you think "real mathematicians" frown? I can offer one explanation that is relevant: mathematicians often love to ensure that everything they do is completely rigorous. Although rigor alone is not mathematics, it is necessary. For example, a good number of students learn about the change of variables rule in integration without knowing the formal proof of this rule. And certainly many people do not stop and look up the hypotheses of the relevant theorem every time they apply the rule ...
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Amitesh DattaJun 23 '11 at 7:59

... However, I am fairly certain that many mathematicians would frown upon a student applying the change of variables rule all the time without really knowing that there is a theorem (with proof) behind the rule. Unfortunately, this is how one is taught about the rule in schools these days.
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Amitesh DattaJun 23 '11 at 8:01

2 Answers
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Suppose you have the equation
$$f(x) = g(x).$$
You could "multiply both sides by $dx$"
$$f(x) \, dx = g(x) \, dx$$
and then integrate over some interval
$$\int_a^b f(x) \, dx = \int_a^b g(x) \, dx.$$
However, unless you have learned about differential forms, the second equation above is meaningless. This can easily be avoided by simplying integrating both sides of $f(x) = g(x)$ with respect to $x$ just as you learned in calculus.

Now consider the differential equation
$$f(y) \frac{dy}{dx} = g(x).$$
Here, many people will "multiply by $dx$" to get
$$f(y) \, dy = g(x) \, dx$$
and then integrate both sides. Again, without the machinery of differential forms, this statement is meaningless. Here we can avoid this by integrating the original equation with respect to $x$:
$$\int f(y) \frac{dy}{dx} \, dx = \int g(x) \, dx$$
or to use different notation,
$$\int f(y(x))y'(x)\, dx = \int g(x) \, dx.$$
Then as long as certain hypotheses are satisfied, the change-of-variables (substitution) theorem says that this is just
$$\int f(y) \, dy = \int g(x) \, dx$$
so we are back to where the first method lead us.

The point of this is that manipulations involving "multiplying by $dx$" are usually shorthand for some more "rigorous" method. However, if you are aware of (and comfortable with) the mathematics going on behind the scenes, then there is no loss of rigor. "Real mathematicians" don't frown upon using these kinds of manipulations themselves, they just frown upon students using them without knowing why they work.

Great answer so far. What exactly are those "certain hypotheses?"
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trolle3000Jun 23 '11 at 23:43

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If you are referring to the change of variables part, there are a few different COV theorems depending on the integral (Riemann, Lebesgue, etc.) and the types of functions you are dealing with. A weak version for the Riemann integral is: If $\phi\in C^1$ and strictly increasing on $[a,b]$, and if $f$ is integrable on $[\phi(a),\phi(b)]$, then $\int_{\phi(a)}^{\phi(b)} f(t) \, dt = \int_a^b f(\phi(x))\phi'(x) \, dx$.
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CoreyJun 24 '11 at 0:22

Because it is considered as non rigorous. Calculating with infinitesimals is something physicists and mathematicians have done since at least the time of Fermat, even long before that. But it always felt awkward to work with infinitesimals as they seemed to lead to contradictory statements if not manipulated carefully.

In the 19th century, under the impulse of Cauchy, calculus started to be formalized and the calculus of infinitesimals was replaced by the more precise and rigorous $\epsilon-\delta$ methodology. Physicists didn't take much notice however since calculating with differentials still worked and was more practical.

In the 1960's however, Abraham Robinson developed a rigorous approach to manipulating infinitesimals now called non-standard analysis. At the time it was controversial, nowadays, I think it is more accepted. That doesn't mean physicists really care that much, but in a way, Robinson vindicated their approach. Although you still need some work to make things rigorous.

I don't quite see how that's relevant — differential forms have nothing to do with infinitesimals (in modern approach, at least)
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Grigory MJun 23 '11 at 8:33

I don't see how differential forms are involved with the OP? Because he is speaking about integration? Then I guess the theory of differential forms is another approach towards treating expressions with $dx$.
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RaskolnikovJun 23 '11 at 8:37

Even the title of OP mentions differential form ($dx$)!.. Well, frankly speaking, it's not very clear what OP is asking about. But multiplication both sides (of PDE, say) by $dx$ and integrating (almost verbatim from OP) makes perfect sense in the context of differential forms...
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Grigory MJun 23 '11 at 8:43

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OK, he uses $dx$, but doesn't mention differential forms. But I agree that the question makes sense in the context of differential forms. I just thought it also makes sense from the point of view of the theory of infinitesimals, which is how $dx$ was first conceived by Leibniz.
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RaskolnikovJun 23 '11 at 8:48